*công thức : tan x = cot (\(\frac{\pi}{2}\)- x) áp dụng vào làm bạn nhé

a/ \(\tan^2x-\cot^2\left(x-\frac{\pi}{4}\right)=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-1-\frac{1}{\sin^2\left(x-\frac{\pi}{4}\right)}+1=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\sin x.\cos\frac{\pi}{4}-\cos x.\sin\frac{\pi}{4}\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\left(\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x\right)^2}=0\)

\(\Leftrightarrow\frac{1}{\cos^2x}-\frac{1}{\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x}=0\)

\(\Leftrightarrow\frac{1}{2}\sin^2x-\sin x.\cos x+\frac{1}{2}\cos^2x-\cos^2x=0\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}\cos^2x-\sin x.\cos x-\frac{1}{2}\cos^2x=0\)

\(\Leftrightarrow\cos^2x+\sin x.\cos x-\frac{1}{2}=0\)

Đến đây là dễ r nha bn :3

a. ĐKXĐ: ...

Ta có: \(\left\{{}\begin{matrix}VT=\left(tanx-cotx\right)^2+2\ge2\\VP=1+cos^2\left(3x+\frac{\pi}{4}\right)\le2\end{matrix}\right.\)

Dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}tanx-cotx=0\\cos^2\left(3x+\frac{\pi}{4}\right)=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cos2x=0\\sin\left(3x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\pi}{4}+\frac{k\pi}{2}\\x=-\frac{\pi}{12}+\frac{k\pi}{3}\end{matrix}\right.\)

\(\Rightarrow x=\frac{\pi}{4}+k\pi\)

b.

\(\Leftrightarrow\frac{2\pi}{3}\left(sinx-1\right)=k2\pi\)

\(\Leftrightarrow sinx-1=3k\)

\(\Leftrightarrow sinx=3k+1\)

Do \(-1\le sinx\le1\)

\(\Rightarrow-1\le3k+1\le1\Rightarrow-\frac{2}{3}\le k\le0\)

\(\Rightarrow k=0\)

\(\Rightarrow sinx=1\)

\(\Rightarrow x=\frac{\pi}{2}+k2\pi\)

c.

ĐKXĐ: ...

\(\Leftrightarrow\frac{\pi}{4}\left(cosx-1\right)=-\frac{\pi}{4}+k\pi\)

\(\Leftrightarrow cosx-1=4k-1\)

\(\Leftrightarrow cosx=4k\)

Mà \(-1\le cosx\le1\Rightarrow-1\le4k\le1\)

\(\Rightarrow-\frac{1}{4}\le k\le\frac{1}{4}\Rightarrow k=0\)

\(\Rightarrow cosx=0\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(DKXD:\left\{{}\begin{matrix}\cos\left(2x+\frac{\pi}{8}\right)\ne0\\\sin\left(x-\frac{3\pi}{4}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+\frac{\pi}{8}\ne\frac{\pi}{2}+k\pi\\x-\frac{3\pi}{4}\ne k\pi\end{matrix}\right.\)

\(pt\Leftrightarrow\tan\left(2x+\frac{\pi}{8}\right)=-\cot\left(x-\frac{3\pi}{4}\right)=\tan\left(x-\frac{3\pi}{4}+\frac{\pi}{2}\right)\)

\(\Leftrightarrow2x+\frac{\pi}{8}=x-\frac{3\pi}{4}+\frac{\pi}{2}+k\pi\)

\(\Leftrightarrow x=-\frac{3}{8}\pi+k\pi\)

a: \(\Leftrightarrow cos2x=\dfrac{1}{\sqrt{2}}\)

=>2x=pi/4+k2pi hoặc 2x=-pi/4+k2pi

=>x=pi/8+kpi hoặc x=-pi/8+kpi

b: \(\Leftrightarrow sinx=sin\left(\dfrac{pi}{2}-3x\right)\)

=>x=pi/2-3x+k2pi hoặ x=pi/2+3x+k2pi

=>4x=pi/2+k2pi hoặc -2x=pi/2+k2pi

=>x=pi/8+kpi/2 hoặc x=-pi/4-kpi

d: \(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=-sin\left(3x+\dfrac{pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=sin\left(-3x-\dfrac{pi}{4}\right)\)

\(\Leftrightarrow cos\left(x+\dfrac{pi}{3}\right)=cos\left(3x+\dfrac{3}{4}pi\right)\)

=>3x+3/4pi=x+pi/3+k2pi hoặc 3x+3/4pi=-x-pi/3+k2pi

=>2x=-5/12pi+k2pi hoặc 4x=-13/12pi+k2pi

=>x=-5/24pi+kpi hoặc x=-13/48pi+kpi/2

e: \(\Leftrightarrow sinx-\sqrt{3}\cdot cosx=0\)

\(\Leftrightarrow sin\left(x-\dfrac{pi}{3}\right)=0\)

=>x-pi/3=kpi

=>x=kpi+pi/3

\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)

\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)

\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)

1.

\(sin\left(4x-10^0\right)=\dfrac{\sqrt{2}}{2}\)

\(\Leftrightarrow sin\left(4x-10^0\right)=sin45^0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10^0=45^0+k360^0\\4x-10^0=135^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=55^0+k360^0\\4x=145^0+k360^0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=13,75^0+k90^0\\x=36,25^0+k90^0\end{matrix}\right.\) (\(k\in Z\))

2.

Đề không đúng

3.

ĐKXĐ: \(\left\{{}\begin{matrix}cos2x\ne0\\cosx\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(tan2x=tanx\)

\(\Rightarrow2x=x+k\pi\)

\(\Rightarrow x=k\pi\)

4.

\(cot\left(x+\dfrac{\pi}{5}\right)=-1\)

\(\Leftrightarrow x+\dfrac{\pi}{5}=-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=-\dfrac{9\pi}{20}+k\pi\) (\(k\in Z\))

5.

\(cos3x=sin5x\)

\(\Leftrightarrow sin5x=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{\pi}{2}-3x+k2\pi\\5x=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}8x=\dfrac{\pi}{2}+k2\pi\\2x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{4}\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) (\(k\in Z\))